Solving Logarithmic Equations

Example 2.4.1:

Solve the following equations:

(i) log (2x +1) – log (3x – 2) = 1 
(ii) log₈x – 4log₈x = 2
(iii) log x + log (x + 3) = 1 
(iv) log x – log (2x – 1) = 1  

Solution:

(i) log (2x + 1) – log (3x – 2) = 1 

⇒ \( \scriptsize \log \normalsize \frac{2x + 1}{3x – 2} = \scriptsize 1\) 

change to index form

⇒  \( \normalsize \frac{2x + 1}{3x – 2} = \scriptsize 10^1 \) 

cross multiply

⇒ 2x + 1 = 10(3x – 2)

2x + 1 = 30x – 20  

collect like terms

28x = 21

x = \( \frac{21}{28} \)

x = \( \frac{3}{4} \)

(ii) log₈x – 4log₈x = 2

⇒ \( \scriptsize \log_8 x\; – \; \log_8 x^4 = 2 \)

\( \scriptsize \log_8 \left [ \normalsize \frac{x}{x^4} \right] \scriptsize = 2\)

change to index form

i.e.   \( \frac{x}{x^4} = \scriptsize 8^2 \)

⇒ \( \frac{1}{x^3} = \scriptsize 8^2 \)

⇒   \( \scriptsize x^3 = \normalsize \frac{1}{\left(2^3 \right)^2} \)

⇒   \( \scriptsize x^3 = \normalsize \frac{1}{2^6 } \)

⇒ \( \scriptsize x^3 = 2 ^{-6} \)

Take the cube root of both sides

i.e.     \( \scriptsize \sqrt[3]{x^3} = \sqrt[3]{2 ^{-6} }\)

⇒     \( \scriptsize x = 2 ^{-2} \)

⇒     \( \scriptsize x = \normalsize \frac{1}{4}\)

(iii) log x + log (x + 3) = 1 

log [x(x + 3)] = 1

log (x² + 3x) = 1

Change to index form

\( \scriptsize x^2 \: + \: 3x = 10^1 \)

\( \scriptsize x^2 \: + \: 3x \: – \: 10 = 0 \)

(x – 2)(x + 5) = 0

⇒  x = 2 or -5.

(iv) log x – log (2x – 1) = 1

\( \scriptsize \log \normalsize \frac{x}{2x – 1} \scriptsize = 1 \)

change to index form

⇒    \( \normalsize \frac{x}{2x – 1} \scriptsize = 10 \)

cross multiply

i.e.    x = 20x – 10

collect like terms

⇒  19x = 10

x = \( \frac {10}{19} \)

Example 2.4.2:

Given log 2 = 0.3010 and log 3 = 0.4771 and log 7 = 0.8451

Evaluate the following:  

(i) log 8.1                
(ii) log 42            
(iii) log 37.5            
(iv) log √60    
(v) log 0.72 

Solution:

(i) log 8.1

log 8.1 = \( \scriptsize \log \normalsize \frac {81}{10} \)

= log 81 – log 10 

= \( \scriptsize \log 3^4 – 1 \)

= 4log3 – 1

(From the question log 3 = 0.4771, so we substitute)

= 4(0.4771) – 1

= 1.9084 – 1 

= 0.9084

(ii) log 42

= log (7 × 6)

= log 7 + log 6

= log 7 + log (2 × 3)

= log 7 + log 2 + log 3

Given log 2 = 0.3010 and log 3 = 0.4771 and log 7 = 0.8451

= 0.8451 + 0.3010 + 0.4771

= 1.6232

(iii) log 37.5 

\( \scriptsize \log \left (\frac {375}{10} \right) \)

= \( \scriptsize \log \left ( \frac {75}{2} \right) \)

= log 75 – log 2 

= log (25 × 3) – log 2 

= log 25 + log 3 – log 2

= \( \scriptsize \log 5^2 \: + \: \log 3\: – \: \log 2 \)

= 2log 5 + log 3 – log 2 

= \( \scriptsize 2 \left [ \log \frac{10}{2} \right ] \: + \: \scriptsize log3 \: – \: log2\)

= 2[log 10 – log 2] + log 3 – log 2   

= 2[1 – 0.3010] + 0.4771 – 0.3010

= 2[0.699] + 0.4771 – 0.3010

= 1.398 + 0.4771 – 0.3010

= 1.8751 – 0.3010

= 1.5741

(iv) \( \scriptsize \log \sqrt{60} \)

= \( \scriptsize \log 60^{\frac{1}{2}} \)

⇒ \( \frac{1}{2} \scriptsize \log 60 \)

⇒ \( \frac{1}{2} \scriptsize \log \left (6 \; \times \; 10 \right) \)

⇒ \( \frac{1}{2} \left [ \scriptsize \log 6 \; + \; \log 10 \right] \)

⇒ \( \frac{1}{2} \left [ \scriptsize \log \left(2 \; \times \; 3 \right) \; + \; \log 10 \right] \)

⇒ \( \frac{1}{2} \left [ \scriptsize \log 2 \; + \; \log 3 \; +\; 1\right] \)

⇒ \( \frac{1}{2} \left [ \scriptsize 0.3010 \; + \; 0.4771 \; +\; 1 \right ] \)

⇒ \( \frac{1}{2} \scriptsize [1.7781] \)

⇒ 0.88905

= 0.8891 (4 s.f)

(v) log 0.72

⇒ \( \scriptsize \log\frac {72}{100}\)

⇒ log 72 – log 100 

⇒ \( \scriptsize \log (9 \; \times \; 8) \; – \; \log 10^2 \)

⇒ log 9 + log 8 – 2 

⇒ \( \scriptsize \log 3^2\; + \; \log 2^3 \; – \; 2 \)

⇒ 2log 3 + 3 log 2 – 2 

⇒ 2(0.4771) + 3(0.3010) – 2

⇒ 0.9542 + 0.9030 – 2

⇒ 1.8572 – 2

⇒ -0.1428

Example 2.4.3:

Solve the following Simultaneous equations:

(i) log₆x – log₆y = 1,    log₆x + log₆y = 3
(ii) log₂x + log₂y = 5,    log₄x – log₄y = -0.5

Solution (i):

1^st Method:

(i). log₆x – log₆y = 1 …….. (i)

(ii). log₆x + log₆y = 3 ……..(ii)

add equations (i) and (ii) 

⇒  2log₆x = 4

⇒  log₆x = 2

change to index form

⇒ 6² = x

⇒ x = 36

substituting for x in equation (ii)

⇒ log₆36 + log₆y  = 3

⇒ log₆6² + log₆y  = 3

⇒ 2log₆6 + log₆y  = 3

⇒ 2 x 1 + log₆y = 3

⇒ 2 + log₆y = 3

log₆y = 1

⇒ y = 6¹

y = 6

2^nd Method 

(i) log₆x – log₆y = 1 …….. (i)

(ii) log₆x + log₆y = 3 ……..(ii)

using the laws of logarithm for equations (i) and (ii)

⇒ \( \scriptsize log_6 \normalsize \frac{x}{y} \scriptsize = 1 \)

⇒ \( \left( \frac{x}{y} \right) = \scriptsize 6^1 \)

⇒ x = 6y      ………(iii)

also, log₆ xy = 3

⇒ xy = 6³ ………..(iv)

substitute for x in equation (iv)

⇒ (6y)y = 216

⇒ 6y² = 216

⇒ y² = 36

⇒ \( \scriptsize y = \pm 6 \)

substitute for y in equation (iii)

when y = -6, x = -36, y = 6, x = 36

Therefore, \( \scriptsize x = \pm 36 \)

Solution (ii):

(i). log₂x + log₂y = 5 …………..(i) 

(ii) log₄x – log₄y = -0.5 …………..(ii)

change equation (ii) to base 2

⇒ \( \frac{\log_2 x}{\log_2 4} \; – \; \frac{\log_2 y}{2 \log_2 2} = \scriptsize \; -0.5 \)

⇒ \( \frac{\log_2 x}{2 \log_2 2} \; – \; \frac{\log_2 y}{2 \log_2 2} = \; -\frac{1}{2} \)

⇒ \( \scriptsize \frac{1}{2} \log_2 x \: – \: \frac{1}{2} \log_2 y = \: – \: \frac{1}{2} \)

divide through by \( \frac{1}{2} \)

⇒ \( \scriptsize \log_2 x \: – \: log_2 y = \: -1 \) ……………..(iii)

⇒ \( \scriptsize \log_2 x \: – \: log_2 y = 5\) ……………..(i)

add equations (i) and (iii) 

2 log₂x = 4

log₂x = 2

x = 2²

x = 4

substitute for x in equation (i)

⇒ log₂4 + log₂y = 5

log₂4y = 5

4y = 2⁵

4y = 32

⇒ y = \( \frac{32}{4} \)

⇒ y = 8